Brent,

But the question doesn't call for a combination. If you have already identified BG as a possibility, then adding GB is creating something different.

Rastis wrote:Brent,

But the question doesn't call for a combination. If you have already identified BG as a possibility, then adding GB is creating something different.

OK, but if you limit this to BB, GB and GG, then you have to acknowledge that the GB combination is twice as likely to occur relative to either the BB or GG combinations. In other words, 50% vs 25% for either.

So with a girl being at least one of the children, the choices are limited to BG and GG, at 50% and 25% likelihood respectively. Therefore, the GG combination comprises 25% of the 75% probability of either occurring, or 1/3.

Thank, you, Ceilidh! Your explanation is exactly what I was trying to convey.

This question is indeed too advanced for the GMAT. It actually tests an abstract concept in Probability sometimes known as the Monty Hall problem (from Lets Make a Deal). It goes like this: You're on Lets Make a Deal and are asked to choose one out of three curtains, one which has a car behind it, the other two have goats. Suppose you choose Curtain #1. Monty then tells you that Curtain #2 has a goat behind it and that you may switch your choice to Curtain #3. The question: should you choose Curtain #3 or would it make no difference whether you stay with Curtain #1? The answer: you should switch, because then theoretically you would have a 2/3 of chance of winning versus only 1/2. But here's where things get really bizarre (which is why it wouldn't be on the GMAT), long-term random selections yield roughly 70% for Curtain #3 with the car and 30% for Curtain #1. But the math works out to 66.67% and 33.33%. It has something to do with the laws of large numbers, a concept certainly not on the GMAT. Just sayin'.

I concluded that the probability is half because order here doesn't matter. isn't B-G the same as G-B?

Brent@GMATPrepNow wrote:

Rastis wrote:A couple has two children, one of whom is a girl. If the probability of having a girl or a boy is 50%, what is the probability that the couple has two daughters?

a) 1/8
b) 1/4
c) 1/3
d) 1/2
e) 2/3

If we list the possibilities in the form "1st child - 2nd child," the 4 possible cases are:
B-B
B-G
G-B
G-G

We're told that at least one of the children is a girl. In other words, the couple has one of these 3 EQUALLY-LIKELY cases:
B-G
G-B
G-G

The question asks for the probability of G-G, so the probability = [spoiler]1/3[/spoiler]

Answer: C

Cheers,
Brent

I concluded that the probability is half because order here doesn't matter. isn't B-G the same as G-B?

Brent@GMATPrepNow wrote:

Rastis wrote:A couple has two children, one of whom is a girl. If the probability of having a girl or a boy is 50%, what is the probability that the couple has two daughters?

a) 1/8
b) 1/4
c) 1/3
d) 1/2
e) 2/3

If we list the possibilities in the form "1st child - 2nd child," the 4 possible cases are:
B-B
B-G
G-B
G-G

We're told that at least one of the children is a girl. In other words, the couple has one of these 3 EQUALLY-LIKELY cases:
B-G
G-B
G-G

The question asks for the probability of G-G, so the probability = [spoiler]1/3[/spoiler]

Answer: C

Cheers,
Brent